Question

How do i balance this equation ?

Answer 1

Complete combustion of butane........? Well the products are ALWAYS `"carbon dioxide"` and `"water"`.

Explanation:

And the normal rigmarole is to balance the carbons, as carbon dioxide, and the hydrogens as water, and then balance the oxygens......

`C_4H_10(g) +O_2(g) rarr 4CO_2(g) +H_2O(l)` `"carbons balanced"`

`C_4H_10(g) +O_2(g) rarr 4CO_2(g) +5H_2O(l)` `"hydrogens balanced"`

`C_4H_10(g) +13/2O_2(g) rarr 4CO_2(g) +5H_2O(l)` `"oxygens balanced"`

If you like you could double the entire equation to remove the half-integral coefficient in front of the dioxygen, i.e.

`2C_4H_10(g) +13O_2(g) rarr 8CO_2(g) +10H_2O(l)`

I have never seen the need to do so, in that I think the arithmetic is easier in the former case. And while I cannot have 1/2 a dioxygen molecule, I can certainly have half a mole of dioxygen molecules. It is up to you (and your prof) which you prefer. If we double the reaction, note that we must necessarily double the thermodynamic properties.

Note that this reaction follows the pattern of all stoichiometric equations. Garbage out equals garbage in, and MASS is conserved, and charge is conserved.

Can you do the corresponding combustion for `"pentane"`, `C_5H_12`? Try it out!